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\title{\ifnum\TeacherEdition=1 TEACHER VERSION \else STUDENT VERSION \fi\\Modeling Car Suspensions}





\markboth{Car Suspensions}{Car Suspensions}


\author{
\begin{tabular}{ccc}
Therese Shelton  &\quad\quad &   Brian Winkel\\
Southwestern University    &\quad \quad&   Director SIMIODE \\ 
Georgetown TX 78628 USA    &\quad\quad &       Chardon OH 44024 USA\\
shelton@southwestern.edu     &\quad \quad&     BrianWinkel@simiode.org \\
\end{tabular}
}

\makeStitlePDFLaTex


\begin{abstract}
We examine the spring-mass-dashpot that is part of a car suspension, how the ride is related to parameter values, and the effect of changing the angle of installation. We model a ``quarter car'', meaning a single wheel.
\end{abstract}


\section*{SCENARIO DESCRIPTION}
Some of us may be familiar with car suspensions, others of us clueless, and many of us in between.  We can use the interplay between math/physics and cars to improve understanding of all by examining different combinations of suspension components. 
A car suspension acts as a spring-mass-dashpot system on each wheel. This allows the car to stay on the road and enable good handling of the car under normal driving conditions.  We examine a ``quarter car'' model that involves a second order differential equation for one wheel.  
Students might refer to other sources about solutions to a second order, linear, homogeneous differential equation with constant coefficients. 
 
  
\section*{GENERAL MODEL:  Relating Car Suspensions and Spring-Mass-Dashpot}

A car suspension is reasonably complicated, with multiple parts and various types, and has evolved over at least five hundred years of human transportation when the passenger part of a horse-drawn carriage was ``slung from leather straps attached to four posts of a chassis that looked like an upturned table. Because the carriage body was suspended from the chassis, the system came to be known as a `suspension' - a term still used today.'' \cite{HowStuffWorks}  
Today, a car suspension includes coil springs with shock absorbers, and more, as in Figure  \ref{fig:Fig1}(a). 
(For instance, consider \cite{KnowParts}, \cite{HowStuffWorks}, \cite{SpringTypes}).

A stylized version of an uninstalled spring assembly is shown in Figure  \ref{fig:Fig1}(b).  A simplified version of the spring and shock absorber that directly relates to our differential equation is shown in Figure  \ref{fig:Fig1}(c). 


 \begin{figure}[h!]
   \includegraphics[scale=.2]{Mercedes-Benz_C111_suspension.PNG} \hspace{0.5in}
   \includegraphics[scale=.2]{SpringMicaPic2.pdf} \hspace{0.75in}
   \includegraphics[scale=.5]{SpringMassDashpot2.pdf} 
   \caption{(a) Suspension  \hspace{0.25in} (b) Spring Assembly   \hspace{0.25in} (c)  Diagram   }   \label{fig:Fig1}  %  reference as  \ref{fig:Fig1}
\end{figure} 

We model the vertical displacement $y(t)$ from its equilibrium position, meaning when the vehicle is at rest.   
The differential equation, Equation  \ref{eq:GenIVP},  for the displacement in a standard spring-mass-dashpot situation shows a sum of the three forces in Figure  \ref{fig:Fig1}(c).  Our righthand side is zero for no external forces, such as for a car on a smooth road.
\begin{equation}\label{eq:GenIVP}
M \cdot y''(t) + c \cdot y'(t) + k \cdot y(t) = 0 \, , \quad y(0) = y_0 \, , \quad  y'(0) = 0 \, .
\end{equation}
Here are the parameters of our system: 

\begin{itemize}
\item $M$ is the mass on one wheel.  Let us consider one quarter of the mass of a car. 
\item $k$ is the spring constant. A spring restores motion proportional to the amount of displacement. Here is a statement of Hooke's Law in a car context:  the more a car spring is stretched or compressed, the more it acts to return to an equilibrium position of rest. 
\item $c$ is the damping coefficient for the reduction in movement by the shock absorber.  Here is a statement of the velocity relationship of damping in a car context:  a shock absorber resists motion more when the suspension moves faster. 
``All modern shock absorbers are velocity-sensitive - the faster the suspension moves, the more resistance the shock absorber provides. This enables shocks to adjust to road conditions and to control all of the unwanted motions that can occur in a moving vehicle, including bounce, sway, brake dive and acceleration squat.'' \cite{HowStuffWorks}   
\item $y(0) = y_0 \, , \quad  y'(0) = 0$.  We begin our clock, $t_0=0$, when the car is at a small displacement, $y_0$, with initial velocity zero. That could be from a person hopping onto the hood of a car and compressing the spring for a negative initial displacement, or at some instant in a lab test.
\end{itemize}

Throughout, we could consider differences between front and rear suspensions. 
For the examples in \cite{SpringRateChart}, the rear axle bears between $48\%$ and $53\%$ of the total car mass.   
Some automotive sites, such as \cite{GenSprings} and \cite{Moog} have specific information about front and rear car springs. 
Front and rear dampers are studied in   \cite{DampingCoefs}.
Students might enjoy a scavenger hunt to gather information on specific vehicles, and some might share their expert knowledge.

\subsection*{Analyze General Model}
\begin{enumerate}
\item State the characteristic equation for Equation  \ref{eq:GenIVP} and its roots. 
\item Why would the suspension system with a shock absorber give a more comfortable ride? State the differential equation if we had no shock absorber. 
\item Relate the mass, damping coefficient, and spring constant to a critically damped situation. 
Recall that ``critical damping'' means the combination of equation parameters that first results in no oscillations. 
\end{enumerate}


 
\section*{SPECIFIC SITUATIONS}

Let our time variable, $t$, be measured in seconds ($s$), and the displacement, $y$ in meters ($m$), with mass $M$ in kilograms ($kg$), so that each of the three forces that are summed in Equation  \ref{eq:GenIVP} is measured in Newtons. (Recall that 1 Newton = 1 $kg \cdot m/s^2$.) 
A compact cross-over sport utility vehicle might be $3,400 \;  lbs$, with a mass on one wheel of $386 \; kg$; a basic full-size pick-up truck might be $5,000 \; lbs$, with a mass on one wheel of $568  \; kg$.
We measure $k$ in Newtons per meter ($N/m$), with reasonable values between $16,000$ and $80,000$ Newtons per meter ($N/m$).
We measure $c$  in $N \cdot s/m$, with reasonable values between  $1,000$ and $21,000$. 

Spring constant values varied a lot in our searches: 
 between $16,000$ and $28,000$ Newtons per meter ($N/m$) on  \cite{Moog}; 
 over $57,000 \; N/m$ in the example in \cite{SpringExample}; 
  $64,000$ and $73,000$ $N/m$  on  \cite{GenSprings}; 
  between $16,000$ and $73,000$ $N/m$  in some portions of  \cite{SpringRateChart}.
 As with most US sites, this required conversions, such as $1\; lb/in\; = \; 175.13  \; N/m$.) 
Values given in \cite{DampingCoefs} for front dampers range between $1,600$ and $21,000$; for rear dampers, between  $870$ and $5,100$.
 

\subsection*{Determine Damping}
We use the following values: 
the mass on one wheel $390 \; kg$, 
  the spring constant $44,000  \; N/m$, 
  and the initial position is $0.2 \; m$ above the static equilibrium.

\begin{enumerate} [resume]
\item  Relate these values to the constants and variables in Equation  \ref{eq:GenIVP}.
\item Determine the values of $c$ for a critically damped system. Graph the resulting displacement. 
\item How long does it take for the critically damped system to stay within a tolerance of $0.1 \; m$ from the static equilibrium?  Illustrate with a graph.
\end{enumerate}

\subsection*{Changing the Ride Quality}
We begin with the following values: 
the mass on one wheel $450 \; kg$, 
  the spring constant $16,000  \; N/m$, 
  the damping coefficient $ 1600  N \cdot s/m$,
    and the initial position is $0.2 \; m$ above the static equilibrium.
\begin{enumerate} [resume] 
\item  
 \begin{enumerate}
    \item Graph the resulting displacement.  
    \item Is this a smooth or bumpy ride?   Is this system underdamped, critically damped, or overdamped?
    \item How long does it take for this system to stay within a tolerance of $0.1 \; m$ from the static equilibrium? Illustrate with a graph.
\end{enumerate}
\item How does the ride change in this situation if the total mass of the car is increased by $1,000 \; lbs$? Show a graph for both rides, and give a brief description.


\item  Consider installing the spring at an angle of $\theta$ from vertical by rotating the top of the spring, as in in Figure  \ref{fig:Fig2}, up to  $\pi/4$ .  
\begin{figure}[h!]
\begin{center}
   \includegraphics[scale=.2]{SpringRotatedPic2.pdf} 
   \caption{Angle of Installation}   \label{fig:Fig2}  %  reference as  \ref{fig:Fig2}
\end{center}
\end{figure}
 \begin{enumerate}
    \item Determine the spring rate correction factor in general, meaning the ratio of the spring constant to equivalent vertical portion. 
    \item Determine the new spring constant $k$ that we would need to compensate for the angles of installation of $\pi/16, \pi/8, \pi/4, $ if we wish an effective spring constant of  $k_v \;= \; 16,000  \; N/m$. 
    \item Compare the rides without additional mass using a spring installed at angles of $20^o$ and  $40^o$. Show a graph for both rides, and give a brief description.
    \end{enumerate}

  \end{enumerate}
  

\subsection*{Conclusion}
\begin{enumerate} [resume] 
\item  Write a short description of how differential equations can model a simple car suspension system.
\end{enumerate}


%%%%%%%%%%%%%%%%%%

\teacher{

\section*{NOTES FOR TEACHERS}
This activity puts the notion of a spring-mass-dashpot system in the context of a simple suspension system in a car. We challenge the students to consider the effects of relating the system components to parameters in the model, and to compare the resulting ride.
See the Mathematica notebook, with accompanying pdf version, in the Supporting Docs with complete analyses.



\section*{ANSWERS}

\subsection*{Analyze General Model}
\begin{enumerate}
\item   The characteristic equation using $y=e^{r t}$ is $M \; r^2 + c \;  r + k = 0$.  \\
Roots are $r \; = \; \frac{1}{2 M} (- \; c  \; \pm \sqrt{c^2-4 \; k \; M}) $
\item The shock absorber provides dampening to the suspension system, so $c=0$ without shock absorbers.  Without dampening, the ride would be bouncy, modeled by the oscillatory solutions $ \cos(\sqrt{k/m} \; t) $,  $\sin( \sqrt{k/m} \; t)$ with constant amplitude. Note that in reality other dampening forces would preclude perpetual motion. 
\item  For a critically damped situation.  $c^2 \; - \; 4  \; k  \; M=0, r = - c/(2M)$.  
\end{enumerate}

\subsection*{Determine Damping}
\begin{enumerate}[resume]
\item   $M=390 \; kg$,  and  $k=44,000  \; N/m$
\item critical damping occurs for $c=2 \sqrt{M \; k} = 8284.93 \; N \cdot s/m$. This is quite a smooth ride, as seen in Figure  \ref{fig:Fig3}(a).
\begin{figure}[h!]
   \includegraphics[scale=.5]{CriticalDampingLbl.pdf} \hspace{.75 in} 
   \includegraphics[scale=.5]{CriticalDampingToleranceLbl.pdf}
    \caption{ (a) Critically Damped \hspace{1.5 in} (b) Tolerance}
     \label{fig:Fig3}  %  reference as  \ref{fig:Fig3}
 \end{figure}  
 \item Because our solution is decreasing from a positive displacement, to stay within a tolerance of $0.1 \; m$ from the static equilibrium we find the first time:  
 $ 2.12434 e^{-10.6217 t} (0.0941469 + 1. t) = 0.1$, so $t=0.158011$ seconds. 
 Figure  \ref{fig:Fig3}(b) illustrates this.
 This is faster than the blink of an eye, which is $1/3$ of a second. 
\end{enumerate}



\subsection*{Changing the Ride Quality}

\begin{enumerate} [resume] 
\item  
 \begin{enumerate} % a,b,c
    \item Figure  \ref{fig:Fig4}(a) shows the resulting displacement.  
\begin{figure}[h!]
   \includegraphics[scale=.5]{UnderdampedLbl.pdf} \hspace{.75 in} 
   \includegraphics[scale=.5]{UnderdampedToleranceLbl.pdf}
    \caption{ (a) Underdamped \hspace{1.75 in} (b) Tolerance}
     \label{fig:Fig4}  %  reference as  \ref{fig:Fig4}
 \end{figure}  
  \item  This underdamped ride is more bumpy than the previous, critically damped, one but would barely be felt, if the oscillations are felt at all. \\
Evidence of underdamping:  The image shows decreasing oscillations, the form of the differential equation solution (see below)  indicates decreasing oscillations, and the discriminant of the characteristic equation is slightly negative ($c^2 - 4*k*m = -2.624 \; 10^7)$.
    \item Because our system oscillates, we check both above and below equilibrium; in this instance, to stay within a tolerance of $0.1 \; m$ from the static equilibrium we still find the time to cross $+0.1$:  
 $ 0.2 \; e^{-1.77778 t} \; \cos(5.69167 t) + 
 0.0624695 \; e^{-1.77778 t} \; \sin(5.69167 t) = 0.1 $, so $t=0.1980971$ seconds. 
  Figure  \ref{fig:Fig4}(b) illustrates this. 
  This system takes a bit longer to reach the tolerance than the previous set-up, but it is still faster than the blink of an eye, which is $1/3$ of a second.
 \end{enumerate} % a,b,c
 
 \item For the new situation with the total mass of the car increased by $1000 \; lbs$,  we solve Equation  \ref{eq:GenIVP2}  with appropriate values.
\begin{equation}\label{eq:GenIVP2}
(M \; + \; m_2) \cdot y''(t) + c \cdot y'(t) + k \cdot y(t) = 0 \, , \quad y(0) = y_0 \, , \quad  y'(0) = 0 \, .
\end{equation}
The addition of $113.636 \; kg$ per wheel makes a slight difference, as seen in  Figure  \ref{fig:Fig5}(a). The oscillations are a bit larger and last slightly longer from the extra mass.  The smoother ride without the extra mass makes sense. 
\begin{figure}[h!]
\begin{center}
   \includegraphics[scale=.5]{CompareMassesLbl.pdf}  \hspace{.75 in} 
   \includegraphics[scale=.5]{CompareAnglesLbl.pdf} 
    \caption{ (a) Added Mass is Dashed  \hspace{1.75 in} (b) Larger Angle is Dashed}
     \label{fig:Fig5}  %  reference as  \ref{fig:Fig5}
\end{center}
 \end{figure}  
 
  \item  
\begin{enumerate}  % a,b,c 
\item   From Figure  \ref{fig:Fig2}, 
we see that the spring rate correction factor is  $\cos(\theta) = (k_v \; y)/(k \; y) = k_v/ k$.
\item Thus, the new spring constant for the installed spring is $k = k_v /  \cos(\theta)$, in general, and  $k = 16 000 /  \cos(\theta)$ for our desired effective spring. Values are given in Table  \ref{fig:Tab1}.

\begin{table}[h!]
\begin{centering}
\begin{tabular}  { | l  | l  | l   | }  
\hline
$\theta$ & $\theta$ &  $k$ \\
\hline
$\pi/16$ & 0.0196  &   16003.1\\
$\pi/8$   & 0.0393  &   16027.8\\
$\pi/4$   & 0.0785  &   16049.5\\
\hline
\end{tabular}
   \caption{Angle and Required Spring Rate}
    \label{fig:Tab1} % reference as   \ref{fig:Tab1}
\end{centering}
\end{table}

    \item   As we see in  Figure  \ref{fig:Fig5}(b), the oscillations for a larger installation angle are a bit larger and last  longer.   The smoother ride with the more effective spring makes sense.  At an angle of $\pi/9$, the effective spring constant is $15,035  \; N/m$, whereas the angle of  $2 \pi/9$ results in an effective spring constant of nearly $12,257  \; N/m$.

\end{enumerate}  % a,b,c

 \end{enumerate}
 }




\begin{thebibliography}{99}

% teacher and student
\bibitem{KnowParts}
Automotive Aftermarket Suppliers Association (AASA), 
``The Online Resource for Today's Technicians''
\url{https://www.knowyourparts.com/technical-resources/}, section ``Suspension and Chassis''
%(a)``Springs and Suspension: Basic Setup,'' 
%\url{https://www.knowyourparts.com/technical-resources/suspension/springs-and-suspenstion/}. 
%defn and brief description spring rate
%image of dusty big coil surrounding a center pole
%(b) ``Complete Suspension Upgrade,'' 
%\url{https://www.knowyourparts.com/technical-resources/suspension/complete-suspension-components-upgrade/}  
%good pic of installed parts, showing two at angles
Accessed 2020, Jun 26. 

% teacher and student
\bibitem{HowStuffWorks}
Harris, William.  (2005) ``How Car Suspensions Work,''
HowStuffWorks.com. 
\url{https://auto.howstuffworks.com/car-suspension.htm} or 
\url{https://auto.howstuffworks.com/car-suspension.htm/printable}
%good images of parts, installed and not.
Accessed 2020, Jul 9.

% teacher and student
\bibitem{SpringTypes}
Meineke,  ``Types of Springs for Your Vehicle,''
\url{https://resource-center.meineke.com/types-of-springs-for-your-vehicle/}
 %good zoomed in pic of installed parts, showing part of one at angle
 %description of spring types
Accessed 2020, Jun 26. 

% teacher and student
\bibitem{Moog}
Moog-Suspension-Parts.com, authorized dealer of Moog parts. 
\url{https://www.moog-suspension-parts.com/universal_coil_springs.asp} 
Accessed 2020, Jun 26. 

% teacher and student
\bibitem{GenSprings}
General Spring of Kansas City, Inc.. 
\url{https://www.generalspringkc.com/coil_springs_and_its_uses_s/2989.htm}
Accessed 2020, Jun 26. 


\bibitem{SpringExample}
Suspension Secrets, ``How to Calculate Coil or Leaf Spring Rates''
\url{https://suspensionsecrets.co.uk/how-to-calculate-coil-or-leaf-spring-rates/}
Accessed 2020, Jun 26.

% teacher and student
\bibitem{DampingCoefs}
Tare, Hitesh K. and Dharankar, C. S.  
``Determination of Damping Coefficient of Automotive Hydraulic Damper Using Sinusoidal Testing,'' 
International Journal of Current Engineering and Scientific Research 3(7), 2016,  
ISSN (print): 2393-8374, (online): 2394-0697.
\url{http://troindia.in/journal/ijcesr/vol3iss7part2/4-13.pdf}


% teacher and student
\bibitem{SpringRateChart}
Three Pedals LLC, ``Spring Rate Chart,''
\url{https://threepedals.com/wp-content/uploads/2018/07/Spring-Rate-Chart-Vehicle-Weights.pdf}. Accessed 2020, Jun 26.

% teacher and student
\bibitem{SpringMassDiagram} 
Wikimedia, author pbroks13, (2008)``Mass spring damper.svg''.  
 \url{https://upload.wikimedia.org/wikipedia/commons/4/45/Mass_spring_damper.svg},
used under public domain, to obtain  Figure  \ref{fig:Fig1}(c).
 Accessed 2020, Jun 26.
 %(image (a) used under public domain from Wikimedia, author pbroks13)  
%  \url{https://upload.wikimedia.org/wikipedia/commons/4/45/Mass_spring_damper.svg}.   1,280 � 996 pixels  24 May 2008

% teacher and student
\bibitem{SuspensionLayout} 
Wikimedia, author Simiprof, (2014) ``Suspension layout Mercedes-Benz C111''. 
 \url{https://upload.wikimedia.org/wikipedia/commons/1/14/Mercedes-Benz_C111_suspension.PNG},
used under public domain, for  Figure  \ref{fig:Fig1}(a).
%(image (c) used under public domain from Wikimedia, author Simiprof)  
% 830 � 565 pixels  Title: Suspension layout Mercedes-Benz C111. Date 	5 September 2014


% teacher and student
\bibitem{WinkelSpringCost}
Winkel, Brian (2020), ``3-031-S-SpringCost''  \teacher{and ``3-031-T-SpringCost''}.

\end{thebibliography}

%}


\end{document}
